what regex expression in python for this text

Question

I have a complicated text file, here is part of it:

& \multirow{2}{*}{52.7} & \multirow{2}{*}{3} & \multirow{2}{*}{$<$ 0.001}\\

I'm interested in the numbers after the {*}. Here is what I tried with no luck:

import re
m = re.findall(r'{\*}{(.+)}', '& \multirow{2}{*}{52.7} & \multirow{2}{*}{3} & \multirow{2}{*} $<$ 0.001}\\')

However, I get the following result:

['52.7} & \\multirow{2}{*}{3} & \\multirow{2}{*}{$<$ 0.001']

I tried many other combinations but I either get the first number (e.g 52.7), or the middle number (3) or the above. How can I get 52.7, 3, $<$ 0.001 in a group (list).

Your first regex obviously won't work (quantifiers are greedy), however I am curious as to what you expect your regex to match in your third example, ie \multirow{2}{*}{$<$ 0.001}? — fge
– fge, Commented Dec 15, 2012 at 23:32
I wanted the bit between the last set of curly brackets; $<$ 0.001. It's working now after adding the ?. — abdulhaq-e
– abdulhaq-e, Commented Dec 15, 2012 at 23:42

ualinker · Accepted Answer · 2012-12-15 23:34:01Z

3

That's because by default + and * operators are greedy. Use non-greedy modification instead:

{\*}{(.+?)}

Reference: http://www.regular-expressions.info/repeat.html ("Watch Out for The Greediness!" section)

edited Dec 15, 2012 at 23:34

answered Dec 15, 2012 at 23:28

ualinker

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Justin McDonald · Accepted Answer · 2012-12-15 23:28:52Z

1

use the following regex expression:

\{\*\}\{(.*?)\}

you should escape all special characters with backslash \ and use non-greedy wildcard .*? in a subclass for result set.

answered Dec 15, 2012 at 23:28

Justin McDonald

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ssedano · Accepted Answer · 2012-12-15 23:41:03Z

1

m = re.findall(r'({\*}{([\d|\.?]+?)})+', '& \multirow{2}{*}{52.7} & \multirow{2}{*}{3} & \multirow{2}{*} $<$ 0.001}\\')
[('{*}{52.7}', '52.7'), ('{*}{3}', '3')]

m = re.findall(r'{\*}{([\d|\.?]+?)}+', '& \multirow{2}{*}{52.7} & \multirow{2}{*}{3} & \multirow{2}{*} $<$ 0.001}\\')
['52.7', '3']

m = re.findall(r'{\*}{(.*?)}', '& \multirow{2}{*}{52.7} & \multirow{2}{*}{3} & \multirow{2}{*} $<$ 0.001}\\')
['52.7', '3', '$<$ 0.001']

edited Dec 15, 2012 at 23:41

answered Dec 15, 2012 at 23:30

ssedano

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